{:[" Find "(dy)/(dx)" and "(d^(2)y)/(dx^(2))],[x=1+t^(2)quad y=t-t^(3)]:}

Differentiate x with respect to t: Differentiate x with respect to t to find dx/dt. x = 1 + t^2 dx/dt = d/dt(1 + t^2) dx/dt = 0 + 2tDifferentiate y with respect to t: Differentiate y with respect to t to find dy/dt. y = t - t^3 dy/dt = d/dt(t - t^3) dy/dt = 1 - 3t^2Find dy/dx: Find dy/dx by dividing dy/dt by dx/dt. dy/dx = (dy/dt) / (dx/dt) dy/dx = (1 - 3t^2) / (2t)Differentiate dy/dx with respect to t: Differentiate dy/dx with respect to t to find d^2y/dx^2. d^2y/dx^2 = d/dt(dy/dx) / (dx/dt) d^2y/dx^2 = d/dt((1 - 3t^2) / (2t)) / (2t)Apply the quotient rule: Apply the quotient rule to differentiate (1 - 3t^2) / (2t). Let u = 1 - 3t^2 and v = 2t. Then du/dt = -6t and dv/dt = 2. Using the quotient rule: (du/dt * v - u * dv/dt) / v^2 d^2y/dx^2 = ((-6t * 2t) - (1 - 3t^2) * 2) / (2t)^2Simplify the expression: Simplify the expression for d^2y/dx^2. d^2y/dx^2 = (-12t^2 - 2 + 6t^2) / (4t^2) d^2y/dx^2 = (-6t^2 - 2) / (4t^2) d^2y/dx^2 = -3/2 - 1/(2t^2)

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Find $\frac{dy}{dx}$ and $\frac{d^2y}{dx^2}$ , $x=1+t^2\quad y=t-t^3$

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Calculus

Find derivatives of sine and cosine functions

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Q. Find

\frac{dy}{dx}

and

\frac{d^2y}{dx^2}

x=1+t^2\quad y=t-t^3

Differentiate $x$ with respect to $t$ : Differentiate $x$ with respect to $t$ to find $\frac{dx}{dt}$ . $\newline$ $x = 1 + t^2$ $\newline$ $\frac{dx}{dt} = \frac{d}{dt}(1 + t^2)$ $\newline$ $\frac{dx}{dt} = 0 + 2t$
Differentiate $y$ with respect to $t$ : Differentiate $y$ with respect to $t$ to find $\frac{dy}{dt}$ .
$y = t - t^3$
$\frac{dy}{dt} = \frac{d}{dt}(t - t^3)$
$\frac{dy}{dt} = 1 - 3t^2$
Find $\frac{dy}{dx}$ : Find $\frac{dy}{dx}$ by dividing $\frac{dy}{dt}$ by $\frac{dx}{dt}$ .
$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$
$\frac{dy}{dx} = \frac{1 - 3t^2}{2t}$
Differentiate $\frac{dy}{dx}$ with respect to $t$ : Differentiate $\frac{dy}{dx}$ with respect to $t$ to find $\frac{d^2y}{dx^2}$ .
$\frac{d^2y}{dx^2} = \frac{d}{dt}\left(\frac{dy}{dx}\right) / \left(\frac{dx}{dt}\right)$
$\frac{d^2y}{dx^2} = \frac{d}{dt}\left(\frac{1 - 3t^2}{2t}\right) / \left(2t\right)$
Apply the quotient rule: Apply the quotient rule to differentiate $(1 - 3t^2) / (2t)$ . $\newline$ Let $u = 1 - 3t^2$ and $v = 2t$ . $\newline$ Then $\frac{du}{dt} = -6t$ and $\frac{dv}{dt} = 2$ . $\newline$ Using the quotient rule: $(\frac{du}{dt} \cdot v - u \cdot \frac{dv}{dt}) / v^2$ $\newline$ $\frac{d^2y}{dx^2} = ((-6t \cdot 2t) - (1 - 3t^2) \cdot 2) / (2t)^2$
Simplify the expression: Simplify the expression for $d^2y/dx^2$ .
$d^2y/dx^2 = (-12t^2 - 2 + 6t^2) / (4t^2)$
$d^2y/dx^2 = (-6t^2 - 2) / (4t^2)$
$d^2y/dx^2 = -\frac{3}{2} - \frac{1}{2t^2}$